Week 7: Lecture 35: Navier stokes equation solution for laminar flow through circular pipe

[music] [bell] [music] [bell] [music] Hello everyone. Uh until now in case of pipe flows we [clears throat] looked at very two specific cases that is uh constant wall heat flux case and constant wall temperature case. We looked at the energy balance. Basic energy balance was done. Okay. And then we obtained the temperature distribution of the fluid T bulk mean as a function of length for both constant wall heat flux and constant wall temperature case. Constant wall heat flux and constant wall temperature case. In one of the cases the temperature distribution was linear and T- wall was this way. In the constant wall temperature case T- wall of course was constant and T-bulk mean was varying this way. And we also understood prior to this the concept of uh thermally fully developed. So thermally fully developed flow unlike hydrodnamics where du by dz is equal to zero. Here we have an additional complexity or du by dx d by dx of t surface x - t local x r divided by t surface x minus t bulk mean of x. This is equal to zero. And so [clears throat] thermally fully developed concept concept of bulk mean temperature bulk mean or mixing cup temperature which is a function of x and it is basically obtained by integration of the temperature profile and doing an energy balance. So if this is your temperature profile here we are saying that this is going to be a this is tbulk mean and this is obtained by energy balance. So m dot cp tb bulk mean is equal to integral row u of r t of r comma x into da into of course cp row cp row a u is mass flow rate cp * t is this one integrated over the entire surface area. So this G gives me the average or bulk mean temperature. And then I also understood the idea of thermally fully developed flow. And we also said for thermally fully developed part H is going to be independent of X. So you have a region beyond which the flow is thermally fully developed and for that the heat transfer coefficient is constant. In the developing region the heat transfer coefficient varies this way. This is developing and this is fully developed. This is fully developed case. So these were the things that were done in the uh last couple of modules. Now this is giving you bulk mean temperature, wall temperature etc. But we are interested in heat transfer coefficient. So h or heat transfer coefficient is what is of relevance to us. And we know by definition h is equal to minus k dt by dy at y =0 divided by some reference temperature difference. Here we would say t wall of x minus t bulk mean of x. In external flow, the same definition was given to me as minus k dt by dy at y equal to 0 divided by t wall minus t infinity. Some reference temperature difference was what what was used here. The reference temperature difference is changing because the role of t infinity is now replaced by t bulk mean in case of uh internal flow. But the concept of h being a gradient at the wall measure which is nothing but a fun this is nothing but qp prime. So this one is nothing but qp prime. Q prime is equal to h delta t. The the equation for h or the definition for h is coming from there. Okay. So that that remains the same. Which means what do I need? If I'm looking at internal flow or flow through a pipe, I need to obtain T wall as a function of X, T bulk mean as a function of X and and obviously I need DT by DY or DT by DR. Temperature gradient at the wall evaluated at R= capital R. This is Y =0. Once I get these three quantities, I substitute it in this expression and I will be able to get h and from h I will be able to get my nasult number. n u d is equal to h into d / k of fluid. This is my definition of nasult number. This is obtained from your non-dimensionalization of the energy equation. Okay. So this H bar or the heat transfer coefficient that we are talking about will been given by this definition of H and this into diameter of the pipe divided by KF will give me the nusle number. As an engineer I am interested in this particular quantity. That means the for from a procedural point of view having obtained a temperature distribution of the fluid. How do I get the temperature distribution? I have to solve the energy equation. So the energy equation is a partial differential equation and in in the energy equation I have to solve that will give me t of r comma x. From that I will calculate these particular quantities of interest. Once I calculate this I will use it to get my h and from that the nasult number. So the procedure being energy equation. What is the energy equation? Row cp u dt by dx + v dt by dy is equal to k into d² t by dx 2 + d² t by d y². This is of course in cartisian coordinates. Uh but we would have a similar expression in your cylindrical co coordinates which is for a pipe plus the viscous dissipation term volutric heat generation term everything all those things are there. We are saying that when these are all neglected when these are neglected or very small compared to these. I have a partial differential equation here. This is the conduction term. This is the advection term. Solution of this differential equation will give me temperature distribution. But temperature distribution doesn't fall straight away. What is here? Here you have U and V velocities. This U and V velocities are going to come from your Navia Stokes equation. So whatever we studied in fluid mechanics in pipe flow, correct? That set of equations would be solved to give you U and V and from that you would be able to get the uh solution of the energy equation. That means you have continuity equation. You have R momentum and Z momentum equation. We are saying theta momentum we are neglecting because flow is axis symmetric. Let me write whatever and then you have the energy equation. What are the unknowns? unknowns will be u v uh I'll not call u vw radial component of velocity azimutal component of velocity axial component of velocity pressure and temperature five equations 1 2 3 4 5 equations five unknowns the system is a closed system. So we can technically be able to get a solution for this. But life is not so easy because in reality these are all partial differential equations. So only when we are able to get uh simplified versions of the same we can get analytical solutions. Otherwise most of the solutions would end up being numerical solutions. there you have to go to the computer and then program these equations and get the particular solution. So what we are going to do now in this module and probably the entire next module we are going to look at solution of the combination of these equations. So I will go through the continuity and the r thetaz momentum equations. From that we will get the solution for fluid velocity u of zed actual component of the velocity. That means as the flow is progressing along the length of the pipe how is the velocity varying u z as a function of r or ux. Sorry I keep using zed instead of x but it's the same. So let me write ux. So x momentum or zed momentum whatever you want to call this is ux you can call it. Okay. So this is ux easier for most of you. So get this velocity distribution and once this velocity distribution is obtained use this in the energy equation. Okay, solve this. Get the velocity distribution. Use this in the energy equation. Get the temperature distribution. From this you get the gradient in temperature at the wall. And from this you get your heat transfer coefficient and then the nasalt number. This is what we aim to do. This is possible only for the two specific cases of constant wall heat flux and constant wall temperature case because only for those two cases we are able to get analytical solutions for these. Also we were in [clears throat] fluid mechanics we looked at steady state incompressible [clears throat] laminer flow [cough] negligible viscous dissipation dissipation no volutric heat addition These are from the heat transfer point of view. From the fluid mechanics point of view, if I were to write few more assumptions 1 2 3 4, it would be axis symmetric. Steady would translate to d by dt of any quantity is equal to zero. axis symmetric d by d theta is equal to zero. You can also have no swirl component which means u of theta is equal to zero. We can say parallel flow which means u of r is equal to z. Okay. Then of course in the heat transfer part we are talking of negligible viscous dissipation. No volutric heat addition. You can also say constant properties. All these assumptions are going to be put together in our derivation. So let us take a look at pipe flow solution which has been done in fluid mechanics. So we are going to go through this very fast. I'm not going to write this but we will we'll have it through the slides. Once we go to the energy equation which is new which is as a part of this course we will write it step by step so that you follow the procedure and then you are able to get the temperature distributions. So let's just go back to the uh set of slides. Okay. So this is what I said mass and momentum equations are solved and then you get your uh [clears throat] uh vr v theta vz and pressure and then you go to go and get the friction factor and the pumping power. Okay. So yeah. Okay. So this is a typical uh geometry which is used for pipes. R theta X coordinate system is shown. Okay. And these are your governing equations of interest. Continuity equation is typically written in a compact form where you take the first and the fourth term together and you will be able to put them together as 1 / r d by dr of vr into r. Okay. When you expand that you get the first and the fourth term. I used u in my nomomenclature. The slides are using v. It's the same. U r u theta ux ur is the radial component of the velocity. U theta is the azimutal component of the velocity and u z is the actual component of the velocity flow along the axis. The governing equations are given to you here. This is the uh r momentum equation. The second one is the theta momentum equation. And this is your uh z direction or x direction momentum equation. Okay. The left hand side is the inertial force everywhere. Right hand side the first term is the pressure force. It's obvious. And the term with new is the viscous force. We have neglected body forces and any other forces. So in my list of assumptions you could also put negligible body forces. Okay. And this is my definition of laplashian del squared. Okay. So now we have to simplify this set of three navia stokes equation and one continuity equation for our laminina steady incompressible flow problem with constant properties. Let's do that. So if I look at my continuity equation, we said UR was zero parallel flow. So the first term drops off, the fourth term drops off. We said nothing varies with respect to theta also u theta is zero. So the second term drops off. What we are left is with dy dx of vx is equal to zero. Which means vx does not vary in the x direction. The velocity profile the act the velocity profile is invariant in the x directions once you are in fully developed flow. It means that we are talking of fully developed flow hydrodnamically fully developed flow. This is the mathematical representation of this. Of course when dvx by dx is zero d² vx by dx² obviously is going to be zero that will be used in the next nia stokes equation. So look at the three uh momentum equations. R momentum equation we said VR was zero because of parall flow everything first term second term third term uh first term is vr therefore it goes to zero this is going to be v theta which is zero and this is dvx by dx which we showed from here was the uh part due to fully developed so this is zero v² theta* r that is also also zero because v theta is zero. Left hand side which is the inertial term is zero. Okay then so this is my laplacian this also goes to zero because vr is zero and v theta is zero. That gives me dp by d r is equal to zero. Let's keep that here. This means dp by d r is zero means pressure is not a function of r. Looking at the theta momentum equation, theta momentum equation because u theta is zero. All the terms on the left hand side the inertial forces drop off. The right hand side again the viscous forces identically drop off. So I am left with dp by d theta is equal to zero which means pressure does not vary in the theta direction. So what did I get? Pressure doesn't vary in the r direction. Pressure doesn't vary in the theta direction which means pressure could vary only in the actual direction if at all it is varying. P is not a function of R. P is not a function of theta. P could at best be a function of X or a constant. Okay, repeating this. P is not a function of R. P is not a function of theta. So at best P could be a function of X or it is going to be a constant. Okay, that's what is implied here. So we look at now the X momentum or Z momentum equation. This is the equation here. dvx by dr into vr. This is zero. Okay. Then v theta is zero. So second term is zero. dvx by dx from continuity fully developed flow. So this term also goes to zero. So the left hand side of the x momentum equation the inertial force component all the three terms are going to be zero. I don't know what to do with the pressure. So we'll keep it this way. And now look at the viscous forces terms. dvx by d r² dv x by d r dv d² vx by d theta² and the fourth term which is d² vx by dx² the fourth term because the first derivative of vx0 the second derivative obviously will be zero so this term drops off this one nothing is a function of theta so this term also drops off and we are left with these two terms which here. So this can be of course combined. This will be written as d by dr d by dr of I mean 1 / r d of r dv x by dr. You can write it that way. This is simple. So let's just go to the next slide. Yeah that's been grouped here. That's been grouped here as shown. That's what is written here. So this is my left hand side and the right hand side of the x momentum equation. Left hand side of the x momentum equation was identically zero. Right hand side was partial derivative dp by dx that becomes a regular derivative. Why does it become a regular derivative? It becomes a regular derivative because p is no longer a function of r and theta we have proved. Therefore this partial derivative becomes a regular derivative. And the second term which is the viscous term is also a partial derivative. Now, how does that become a regular derivative? We'll just uh see in a minute. So, what do I have here? I have minus dp by dz plus mu / r d by dr of r dv z by uh sorry is equal to zero. I'll just open the bracket and see whether I have written this correct. So, dvz by d mu rs cancel plus mu bring the r out as constant. Second derivative will be there. Then bring the dvz by dr out and you'll just have this r. So you would get mu into d² vz by d r² + 1 / r dvz by dr. So let's just check whether we had that right. Yeah. So dv d² vx by d r² + 1 / r dvx by d r that's what I have written there. So this one is a partial derivative. Okay. So what I'm trying to say is this one is now a function of zed only. P is a function of zed alone. Therefore dp by dz becomes equal to dp by dz. So this gets replaced. So what I'm saying dp by dz therefore is equal to mu / r d by d of r dvz by d r. Okay. Now this is a function of zed only. This is a function of r only. Correct? So if this is a function of r only then this should become d by dr will become d by dr. Something which is a function of zed is equal to something which is a function of r. That's possible only when this is completely independent of each other and they are constants. So this derivative therefore becomes a regular derivative. And therefore I will write mu / r d by dr of r d vz by dr is equal to dp by dz and I bring I take one take the mu to the other side. R DV Z by D R is equal to 1 / mu DP by DZ and so on and so forth. The maths is done here. So let's take a look at that. So that's what has been done. So I combine this and then I integrate with respect to r. So first integration would be this r² by 2 dp by dx plus a constant a divide through by mu r there is the r here mu is here divide through you'll get r / 2 mu dp by dx plus a by mu r so a by mu can be taken as some b or something doesn't matter [snorts] and this is second integration would give me v x is equal equal to dp by dx into 1 / 2 mu into r² by 2. Uh so you have r² by 2 and you have a two here from 1 / 2 mu it becomes r 2x 4 mu dp by dx plus a by mu log r plus another constant b. So what are the boundary conditions? The boundary conditions are simple whatever we had in pipe flow. First no slip condition at the wall at R equal to capital R U u is vx is zero and second thing at the center line the velocity is maximum. So dvx by dr at r equal to0 is zero. First boundary condition is second boundary condition is this. When you substitute and do the maths here, you will end up with the profile which is given by v of x is minus capital r² by 4 mu dp by dx 1 minus r by r the whole square. This is a paraboid of revolution. It's a parabola but in three dimensions. So it's like a bullet headshaped structure. Paraboid we call that. From this I will be able to get this is my velocity distribution. I can get the maximum velocity. Maximum velocity is at the center. So substitute r=0 you get this one minus capital r² by 4 mud dp by dx that is the maximum velocity. Average velocity when you do the maths you will get it as average velocity u bar or u mean is u max by 2. Volutric flow rate is cross-sectional area times the uh average velocity that's a number. Sheer stress is obtained by the gradient at the wall. So du by dy which will be translated to du by dr. So you have this velocity distribution. Take the gradient with respect to r that's already here. From that substitute r equal to capital r you'll get the shear stress. Skin friction that is nothing but CFX is to all by 12 roaming squared and friction factor is 4 * CFX. So all these things which you saw in fluid mechanics can be obtained. They are given in this set of slides in complete detail. Every little step is given to you. I urge you to take a look at this entire derivation from fluid mechanics point of view. Please write it as you follow through the slides so that you know what simplifications are made, why have they been made and how have they been made. Okay? So that these quantities maximum velocity, average velocity, volutric flow rate, sheer stress, skin friction coefficient, friction factor, all these things have been shown here step by step. Calculate this. go ahead and find out the expressions for everything that you have already seen as a part of a fluid mechanics course. Okay. So once that is done, we are all set with the temp the velocity distribution which is now going to be used in the energy equation. And once the energy equation is recast with the output from the Navia Stokes equation which is the velocity distribution which is shown here which is shown here based on this and the average velocity I will use it in the energy equation and then I will solve for temperature distribution that part in the next module. Thank you. [music] >> [music and bell]